From: mareg@mimosa.csv.warwick.ac.uk ()
Subject: Re: another question of finitely generated group
Date: 8 Nov 2000 22:36:30 GMT
Newsgroups: sci.math
Summary: [missing]

In article <vl4cujkcphga@forum.mathforum.com>,
	maths@21cn.com (zhang xiaolei) writes:
>please help me. i had post this question in severl place but no one
>answare me. here seem to be active, so i repost it here. if someone
>had saw it, i apologize.
>
>suppose there is a finitely generated group G,G contain a subgroup H,
>the index of H is finite(|G:H|<infnity), show that H is also finitely
>generated.(this qusetion is take from "Basic Algebra I" N. Jacobson)

This is standard result, and the proof can be found in many books on
Group Theory, such as the one by M. Hall.

Here is an outline of the proof.

Let G = Hg_1 union Hg_2 union ... Hg_r  be a decomposition into cosets
of H in G. r is finite by assumption. Assume g_1 is the identity element.
For g in G, write r(g) for the element g_i with g in Hg_i.
Let G be generated by finite set X.  Then H is generated by the finite set
Y = { g_i x r(g_i x)^-1 | 1<=i<=r, x in X }.

Proof: Let h in H. Write h as a word x_1 x_2 ... x_n with x_i in X.
(To simplify things, let's assume X is closed under inverses, although that is
not really necessary.) Let r_i = r(x_1 x_2 ... x_i) for 1 <= i <= r.
Then we can write h as a product of elements of Y:
h = (g_1 x_1 r_1^-1) (r_1 x_2 r_2^-1) ... (r_{n-1} x_n r_n^-1)
(note r_n = g_1 = identity as h is in H).

Derek Holt.